# reduce to the case $[0,1]$ in equidistribution modulo $1$

I am trying to prove a theorem using an optimal solution. Given $$(y_n)_{n \in \mathbb{N^{*}}}$$ such that $$\exists \delta > 0, \forall n \in \mathbb{N^{*}} y_{n+1} - y_n \geq \delta$$ then for almost every $$\xi \in \mathbb{R}$$ (Lebesgue) the sequence $$(\xi y_n)_{n \in \mathbb{N^{*}}}$$ is equidistributed modulo $$1$$.

I found a proof considering an arbitrary interval $$[a,b]$$ but some people told me it's sufficient to prove the result for almost every $$\xi \in [0,1]$$ but I am not convinced. In fact, if we prove the result just for almost every $$\xi \in [0,1]$$ the sequence $$\xi y_n$$ does not cover all the interval $$\xi \in [0,1]$$.

I am trying to find a counterexample, if we have a sequence $$\frac{1}{16} \leq y_n \leq \frac{1}{2}$$ then multiplying by $$\xi \in [0,1]$$ gives us $$0 \leq \xi y_n \leq \frac{1}{2}$$ and if we take $$\xi \in [1,2]$$ we got $$0 \leq \xi y_n \leq 1$$. Do you see what I mean? I think that the argument reducing to $$[0,1]$$ is not always true.

I discovered that such an example does not exist. In fact, we have $$y_{n+1} - y_{n} \geq \delta$$ so $$\lim\limits_{n \longrightarrow +\infty}^{}$$. If we suppose that we have proved the result for almost every $$\xi \in [0,1]$$. Then the result is true for almost every $$\xi \in \mathbb{R}$$. In fact, if we take $$a > 0$$, then we got $$a y_{n+1} - a y_{n} \geq a \delta > 0$$ so $$a \xi y_n$$ is equidistributed for almost every $$\xi \in [0,1]$$ then $$\xi y_n$$ is equidistributed for almost every $$\xi \in \mathbb{R}^{+}$$. Then we deduce the result (because if $${x_n}$$ is equidistributed modulo $$1$$ then $${-x_n}$$ is equidistributed modulo $$1$$.